Question

The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

• A. $50m$
• B. $50\sqrt{3}m$
• C. $\frac{50}{3}m$
• D. $\frac{50}{\sqrt{3}}m$

Answer 1

The correct option is C $\frac{50}{3}m$


Let AB be the building and CD be the tower.

Given the angle of elevation of the top of the building from the foot of the tower
$={30}^{\circ }$
and angle elevation of the top of the tower from the foot of the building is
$={60}^{\circ }.$

Height of the tower = 50 m

From the figure, we have
In $△CDB$,

$\frac{CD}{BD}=tan{60}^{\circ }$
$\frac{50}{BD}=\sqrt{3}$
$BD=\frac{50}{\sqrt{3}}---\left(i\right)$

Next in $△ABD$,
$\frac{AB}{BD}=tan{30}^{\circ }$
$\frac{AB}{BD}=\frac{1}{\sqrt{3}}$
$AB=\frac{BD}{\sqrt{3}}$

$AB=\frac{\frac{50}{\sqrt{3}}}{\sqrt{3}}$
$AB=\frac{50}{3}$

Therefore the height of the building
$=\frac{50}{3}m$